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The magnetic field of a cylindrical magn...

The magnetic field of a cylindrical magnet that has a pole face radius `2.8 cm` can be varied sinusoidally between the minimum value `16.8 T` and the maximum value `17.2 T` at a frequency of `50//piHz`. Cross section of the magnetic field created by the magnet is shown in Fig. 3.211. At a radial distance of `2cm` from the axis, find the amplitude of the electric field `(in units of xx 10^(2) mNC^(-1))` induced by the magnetic field variation.

Text Solution

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The correct Answer is:
2
`int vec(E)dvec(l) = -A(dB)/(dt)`
As `B = 17 + (0.2) sin(omegat + phi)`,
`E(2pir) = -pir^(2)(0.2)omega cos (omehat + phi)`
`E = -(r )/(2)(0.2)omega cos (omegat + phi)`
Magnitude of the amplitude `= (R )/(2)(0.2)omega = 2 xx 10^(2)mN//C`
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